Question

Find two rational and irrational numbers between $\sqrt{2}$ and $\sqrt{7}$.

Answer 1

Step 1: Find the rational numbers

Rational numbers are the numbers that can be represented in $\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$q$}\right.$ form where $p$ and $q$ are integers and $q\ne 0$.

Given irrational numbers, $\sqrt{2}$ and $\sqrt{7}$

Let the two rational numbers between $\sqrt{2}$ and $\sqrt{7}$ be $a$ and $b$.

Convert the given irrational numbers into decimals.

We know that, $\sqrt{2}=1.414\dots$ and $\sqrt{7}=2.645\dots$

So, two decimals between $1.414$ and $2.645$ can be $a=1.5$ and $b=2.5$.

These can be written as,

$$\begin{gathered} \Rightarrow a=1.5 \\ a=\frac{15}{10}=\frac{3}{2} \\ \Rightarrow b=2.5 \\ b=\frac{25}{10}=\frac{5}{2} \end{gathered}$$

Step 2: Find the irrational numbers

Rational numbers are the numbers that cannot be represented in $\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$q$}\right.$ form where $p$ and $q$ are integers and $q\ne 0$.

Let the two irrational numbers between $\sqrt{2}$ and $\sqrt{7}$ be $x$ and $y$.

So,

$$\begin{gathered} \Rightarrow x=\frac{1}{2}(\sqrt{2}+\sqrt{7}) \\ x=\frac{(\sqrt{2}+\sqrt{7})}{2} \\ \Rightarrow y=\frac{1}{2}(\sqrt{2}+x) \\ y=\frac{1}{2}(\sqrt{2}+\frac{(\sqrt{2}+\sqrt{7})}{2}) \\ y=\frac{1}{2}\left(\frac{2\sqrt{2}+\sqrt{14}}{2}\right)=\frac{2\sqrt{2}+\sqrt{14}}{4} \end{gathered}$$

Therefore, the two rational number between $\sqrt{2}$ and $\sqrt{7}$ are $\frac{3}{2}$ and $\frac{5}{2}$, while the two irrational numbers are $\frac{(\sqrt{2}+\sqrt{7})}{2}$ and $\frac{2\sqrt{2}+\sqrt{14}}{4}$.